Download JLH – 822 Voltage Switcher Theory of Operation

JLH – 831 Voltage Switcher Theory of Operation
The 831 voltage switch has only one function. It adds fixed amounts of DC voltage to the
incoming control voltage to offset the output (it actually adds current not voltage -- more on
that later). What makes this circuit unique is that it adds these offsets at fixed increments to
provide voltage shift at exact musical intervals when used in conjunction with any 1-volt-peroctave voltage controller oscillator. And, the circuit is configured to be very accurate so that
the keyboard 1-volt-per-octave scale is not compromised.
Let’s examine the circuit. The functions can be broken down into four sections for analysis:

Section # 1 is the interval resistor divider network. This series string of resistors
provides the individually adjustable offset voltages to our “shift” selector switches.

Section # 2 is the summing amp that actually adds the incoming CV with the shift
voltage. A by-product of this summing process is inversion of the voltage.

Section # 3 is a unity-gain inverting amp, which restores the CV to the correct polarity.

Section # 4 is the output driving amp. It is configured to offset the capacitance of cables
and provide a low impedance output capable of driving multiple oscillators without error.
Sections 2, 3, and 4 are duplicated in the “A” and “B” and “C” channels. For the purpose of
discussion, we will examine the A channel, but the same logic applies to the other channels.
First, let’s examine section # 1. Interval shifting is both upward and downward. So, we
need to establish positive and negative voltages to accomplish those shifts. What voltages
do we need? Well, since 1 volt = 1 octave, and there are 12 semitones to an octave, we
need 1/12 of a volt per semitone. The 831 is unique in that you must select your own
shifting intervals at the time of construction. This document will of course use the intervals
I selected for the purpose of illustration. They are:
Shift +5 = up 10 semitones (dom 7th In the octave above)
Shift +4 = up 7 semitones (5th in the octave above)
Shift +3 = up 5 semitones (4th in the octave above)
Shift +2 = up 4 semitones (major 3rd in the octave above)
Shift +1 = up 3 semitones (minor 3rd in the octave above)
Shift 0 = no shift
Shift -1 = down 5 semitones (5th in the octave below)
Shift -2 = down 7 semitones (4th in the octave below)
Shift -3 = down 8 semitones (major 3rd in the octave below)
Shift -4 = down 12 semitones (one octave below)
Shift -5 = down 24 semitones (two octaves below)
= +0.83333 volts
= +0.58333 volts
= +0.41667 volts
= +0.33333 volts
= +0.25 volts
= -0.58333 volts
= -0.41667 volts
= -0.66667 volts
= -1.0 volts
= -2.0 volts
You can use 0.08333 volts per semitone to calculate any shift you want for your 831.
However, shifting is limited to about 3 octaves due to the resistors in the summing amp.
My prototype design featured just these exact voltages delivered to the summing amp for
unity gain. However, after construction of my first 831, I found a problem with that
approach. The load of the summing op amp on the semitone resistor divider network was
causing the voltage to vary slightly through the network when the loading changed (by
rotating the selector switch). Therefore, I decided to change the relationship between the
total R of the network and the load imposed by the summing op amp. My calculations
indicated that changing this by a factor of 4 or 5 would make the undesired voltage
fluctuation so small it would never be noticed. I settled on resistor values that would
produce 5 times the necessary voltage from the semitone resistor divider network.
To calculate all but R1 and R12, let’s assume 1mA of current (we will see later that I actually
use 5mA). To get 0.0833 volts drop across R, I need 83.3 ohms. Remember Ohm’s law
says V=I*R. So, 83.3 ohms X .001 amps = 0.083 volts. Let’s look at the shift UP side first.
For shift +1 to achieve 3 semitones, we need 83.3 X 3 = 250 ohms. Our trimmers are 100
ohms. Ideally, the multiples of 0.083 volts would occur around the center of the trim pots.
So, we must include 50 ohms of VR1 in that value leaving us needing 200 ohms. So R6 =
200 ohms. For shift +2 (4 semitones), we need 83.3 X 4 = 333 ohms. Since we have
accumulated 300 ohms by the time we get through VR5, we need no resistance in the R5
positions. Why is it there you ask? Because, some builders may want to select different
intervals and include a resistor in this location. I wanted a spot on the PCB for it. For shift
+3 (5 semitones), we find the same thing. Resistance should be 83.3 X 5 = 417 ohms. We
need no resistance between VR4 and VR3. In fact, we can see that if we were planning to
select a series of shifts only one semitone apart, we would need smaller value trimmers.
The 822 which includes this feature does indeed use 50 ohm trimmers. For shift +4 (7
semitones), we need 83.3 X 7 = 583 ohms. Since we have accumulated 500 ohms by the
time we get through VR3, we need 33 ohms to put VR2 in the middle of its range. However,
I selected 51 ohms as I found my 100 ohms trimmers were all reading closer to 95 ohms.
Your mileage may vary. Since you know how to calculate the needed resistance, you can
adjust here if you find VR1 and VR2 near the end of their operating range. For shift +5 (10
semitones), we need 83.3 X 10 = 883 ohms. We have accumulated 651 ohms by the time
we get through VR2. We need 182 ohms to put VR1 in the middle of its range. However, I
selected 150 ohms as close enough.
Now, in reality, we need the voltage at these trimmer mid points to be five times what we
calculated. So, we can accomplish that by leaving all the resistor values the same and
causing 5mA of current to flow in this series resistor network. How do we do that? By the
value selected for R1. Remember Ohm’s law? V=I*R also works out to I=V/R or R=V/I. So,
total R determines the current that will flow in this series string. I know my source voltage is
15 volts. Our calculation is R=15/.005. That equals 3000 ohms as the desired total R. You
must subtract out the total resistance of R1 through R6 and VR1 through VR5. So, 3000 –
901 = 2099 ohms. Therefore, 2.1 ohms is perfect. We know it will be slightly different
because of resistor and trimmer tolerance. Anyway, it is close enough and that is why we
have the trimmers.
The same logic applied to understanding the positive half of this voltage divider can be
applied toward understanding the negative half and to selecting your own intervals and
resistor values if you choose to do so.
OK, now it is time to examine the heart of the circuit – section # 2, the summing op-amp:
U1A provides the summing of voltages. This is a very common configuration that will sum
any voltages (currents actually, but that is a lesson on op amps) that are attached to its
inverting (-) input. This configuration does invert the voltage. Why use an inverting amp?
You can configure op amps so they don’t invert. However, it is not an easy task to make a
summing amp that is non-inverting. And, since we are VERY interested in accuracy in this
circuit, the best method is the well-traveled inverting summing amp. The amp has two input
resistors, they are R14 and R16. So, we are summing two voltages – the incoming CV and
the voltage from the shift switch. We also have a feedback resistor (R15) which is used to
set the gain of the amp.
Let’s look first at R14 and R15. Why are they 100K? I could have used many different
values. However, 100K is a good value for R14 to provide high input impedance to the amp
so that the incoming CV is not loaded down. The gain of the amp = R15/ R14. We want the
gain of the CV to be exactly 1. We do not want that critical scaling to change. That is why
R14 and R15 are hand matched to be exactly the same. The 1-volt-per-octave CV passes
through this amp at the exact value it comes in (except inverted).
What about R16? Why is it 499K? Well, remember, I selected resistors so that the voltages
in my resistor divider networks were 5 times what I actually needed. This is where we make
the op amp use that value but divided by 5. The gain of the shift voltage = R15/R16. 100K
/ 499K = 0.2. So, all voltages applied to R16 are divided by 5 (multiplied by 0.2) in the
output, putting them right where we need them. Therefore, the output of U1A, is the sum of
the incoming CV X 1, and the shift voltage X 0.2, but inverted to the negative side of zero.
Let’s look at the other two resistors in this stage. R13 does nothing but establish a
reference to ground for the incoming CV. It should be high enough not to place a load
burden on the proceeding stage. The value is not critical. I usually use about 100K. But,
since we have the A input normalled down to the B input and then to the C input, there are
three in parallel. So, I selected a higher value. What about R17? I selected R17 when this
circuit was configured for R16 to be 100K. Since the change to 499K, I really should have
changed R17. The value is non-critical though, and some designers just connect this to
ground. So, I left it alone as “close enough” But, to be perfect, R17 should be 47K. Should
you change it? Nope. R17 should equal the parallel equivalent of all input resistors plus
the feedback resistor at amp U1A. So, if you calculate that for three 100K resistors you get
33K. That is why I originally selected 33K. For two 100K and one 499K resistor, that value
calculates to around 45K. Hence, my suggestion that a “more correct” value here is 47K.
But, 33K is close enough.
Section # 3 will not take much examination. U3B is an inverting amp. R18 and R19 are
matched so that gain is exactly one. The value can be almost anything. 10K was simply my
choice. All this amp does is restore the output of U1A back to the correct polarity. R20 is
calculated the same as R17. Two 10Ks parallel equals 5K ohm. So, 5K1 was the easy
choice. R21 references the output voltage of U3B to ground for the input of U3A (section #
5). Maybe it could be omitted. But, I like this approach. So, there it is.
Section # 4 (amp U3A) is a non-inverting unity gain op amp. I shamelessly stole this circuit
directly from the MOTM-820 LAG processor. What makes this circuit slightly different is that
it is designed to drive large capacitive loads (like long cables). So, you can drive multiple
oscillators from the single output of the JLH-831 using a multiple. I could not provide
ganged outputs because of space limitations on the 1U panel. It is completely full.
The circuit descriptions for sections 2, 3, and 4 always referenced channel A. However, you
will notice that channel B and C are identical with the exception of the trimmers VR11 and
VR12. Only one interval shift resistor divider network is provided for all three channels. We
want it to have the same effect on each channel. I could have hand matched the resistors
to the exact same size. But, that would have been a HUGE pain because of the differences
between the matched pairs in the channels. But, you will notice that the matched resistors I
supplied with your PCBs were indeed very close in value between channels.
To assure uniform effect on each channel, I inserted trimmers in the B and C channels.
That is why when calibrating the shift trimmers you ALWAYS use the A channel. Trimmers
VR11 and VR12 are used to match the shift in channel B and C to channel A.
Lastly, I will give you a brief explanation of why I increased the current in the resistor divider
networks to 5mA from 1mA. This of course provided five times the voltage I needed, and
then I turned right back around and multiplied that by 0.2 (one-fifth) in the summing amp.
Why you ask? That is a very good question. First, accept that op amps really sum currents,
not voltages. We talk in terms of summing voltages because it is easier for discussion
purpose.
Since the voltage at the wiper of our trim pots determines our shift points, we want that
voltage to remain as constant as possible. What establishes that voltage? The current
through the resistor network of course. So, any current added to this network from the load
of the summing op amp will change our voltage. We don’t want that to happen. How do we
limit that current to an amount that will not be significant enough to effect our divider
network? By increasing the value of R16, 27, and 38 so that this current is limited. Why did
I select 499K? Why not 1 meg? Well remember that the effect of this shifting voltage on the
output of U1A and U1B is the ratio of R15/R16 (looking at the octave switch in channel A.
That equals about 0.2. I know I need -2 volts at the output of this amp for my largest
selected shift (2 octaves down). So, the voltage needed to achieve that output is 5 times
that (or –10). Since my divider network cannot provide voltages in excess of +/- 15 VDC,
five was close to the largest multiplier I could use. Otherwise, we could not achieve the -2
octave selection. I could have gone up to 7 times, but 5 was enough. And, 5 times
provides the opportunity for shifts between 2 and 3 octaves if someone is so inclined.
Please feel free to e-mail me if you have additional questions.
Larry Hendry